Answer
$\dfrac{8}{15}$
Work Step by Step
Let us consider that $I=\iiint_T y^2 dV$
$I= \int_{0}^2 \int_{0}^{2-x} \int_{0}^{2-x-y} y^2 dz dy dx $
Thus, we have $\int_{0}^2 \int_{0}^{2-x}[y^2 z]_{0}^{2-x-y} dy dx=\int_{0}^2 \int_{0}^{2-x}2y^2-xy^2-y^3 dy dx$
and $ \int_{0}^{2}[\dfrac{2y^3}{3}-\dfrac{xy^3}{3}-\dfrac{y^4}{4}]_{0}^{2-x} dx=\int^{0}_{2} \dfrac{(2-x)}{3}-\dfrac{(2-x)^3}{3}-\dfrac{(2-x)^4}{4} dx$
and $\int_0^2\dfrac{(x-2)^4}{12}dx=[\dfrac{(x-2)^5}{(12)(5)}]_0^2$
or, $\iiint_T y^2 dV=[\dfrac{(2-2)^5}{60}-\dfrac{(0-2)^5}{60}]=\dfrac{8}{15}$
